1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108)

1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all

1.	1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108)
Answer» Solution :`underset(1g)(EU)CI_(2)+AgNO_(3) rarr underset(1.29 g)(AGCI)` Since CI atoms are conserved applying POAC for CI atoms moles of CI in `EuCI_(2)` = moles of CI in AgCI `2xx ` moles of `EuCI_(2)=1XX` moles of AgCI `2xx(1)/((x+35.5xx2))=1 xx(1.29)/((108+35.5))` ( where x = atomic weight of Eu ) `:. x= 152.48`